P H Y S 3 7 5 - L A B O R A T O R Y E L E C T R O N I C S I 1 3 . 2 - F O R T N E RDifferentiator and Integrator

Differentiator

An amplifier can utilize the relation between charge and current.

The current is converted to a voltage

For a sinusoidal input vin = V0sint,

The amplitude increases with increasing frequency

IQdtd

------- CVdtd

------= =

vout+

Rf i

i

Ci

vin

vout iRf RfCivind

td----------= =

vind

td---------- V0 tcos=

vout RfCi V0 tcos RfCi vin= =1 of 10

P H Y S 3 7 5 - L A B O R A T O R Y E L E C T R O N I C S I 1 3 . 2 - F O R T N E RIntegrator

The input current can be converted into a charge

Solving for vout:

With a sine wave input

The amplitude decreases with frequency.

vout+Ri

i

Cf

vin

ivinRi------- C f

voutd

td-------------= =

vout1

RiCf----------- vin td K+=

vout1

RiCf---------------vin K+=2 of 10

P H Y S 3 7 5 - L A B O R A T O R Y E L E C T R O N I C S I 1 3 . 2 - F O R T N E RStabilized Integrator

The constant term from the integral is undesirable on the output.

A switch can be used to discharge the capacitor.

The circuit can provide a path for the capacitor to discharge.

The feedback resistor and capacitor have a parallel impedance

The break frequency is f = 1/2RC = 0.007 Hz .

Only DC can feedback for amplification.

vout+Ri = 100 k

iCf = 1 Fvin

Rf = 22 M

ZfRf

1 j+ RfCf---------------------------=3 of 10

P H Y S 3 7 5 - L A B O R A T O R Y E L E C T R O N I C S I 1 3 . 2 - F O R T N E RLimited Differentiator

Differentiator has large amplification at high frequencies.

A high frequency cutoff above the signal range is often needed.

Combine an integrator and differentiator.

The differentiator has a low cutoff

The integrator begins working at

The differentiator will perform well between those two frequencies.

vout+

vin

Ri = 1 k

Rf = 100 kCi = 0.01 F

Cf = 50 pF

vout RfCivind

td----------= fB

12RfCi------------------ 160Hz= =

vout1

RiCf----------- vin td= fB 12RiCf------------------ 3MHz= =4 of 10

P H Y S 3 7 5 - L A B O R A T O R Y E L E C T R O N I C S I 1 3 . 2 - F O R T N E RDC Amplifiers

Logarithmic Amplifier

Use a transistor in the feedback network of an inverting amplifier.

RB compensates for the bias current.

The current Iin is given by

This current must flow into the collector of the transistor

The base-emitter voltage is equal to the negative of Vout

The output depends on the logarithm of the input voltage.

Vout

+

R = 100 k

Vin

RB = 100 k

2N2222

IinVinR--------=

IC I0eVBE VT Vin

R--------= =

Vout V TVinI0R--------log=5 of 10

P H Y S 3 7 5 - L A B O R A T O R Y E L E C T R O N I C S I 1 3 . 2 - F O R T N E RAnalog Product

Combine logarithm amplifiers and summation amplifiers

The final stage is an exponentiator to return the product of the inputs.

(a lnV1 + b)

+

R

V1

RB

+

R

V2

RB

VC = 2b

+

R

R

R

R

(a lnV2 + b)

(a lnV2 + b) + (a lnV1 + b) 2b = a ln(V1V2)

+Vout = cV1V26 of 10

P H Y S 3 7 5 - L A B O R A T O R Y E L E C T R O N I C S I 1 3 . 2 - F O R T N E RCurrent-to-Voltage Converter

Operational feedback relies on current.

This is used for small input signals such as photonic devices.

A parallel feedback capacitor can be used to reduce high frequencies.

The output is usually measured in V/A. The op-amp provides low output impedance, hence higher power.

Vout+

Rf I

I

Vout R fI=7 of 10

P H Y S 3 7 5 - L A B O R A T O R Y E L E C T R O N I C S I 1 3 . 2 - F O R T N E RCurrent-to-Current Converter

An extra resistor to defines the current.

From Kirchoffs Laws:

Combine the equations:

Vout+

Rf Iin

Iin

RL

Rg

IL

Ig

Vout R fIin=

Ig IL I+ in=

Vout RgIg=

IL I in 1RfRg------+

=8 of 10

P H Y S 3 7 5 - L A B O R A T O R Y E L E C T R O N I C S I 1 3 . 2 - F O R T N E RVoltage-to-Current Converter

An op-amp follower can be used to drive a conventional transistorcurrent source.

The current Iout splits through the FET and BJT.

No current passes through the gate of the FET or the v- op-amp input.

All current Iout is present through the resistor IR.

VE = Vin from the op-amp voltage rule.

Iout

+

Rf

I1

I2Vin

R

VE

VS

VG

VC

IR

IoutVinR--------=9 of 10

P H Y S 3 7 5 - L A B O R A T O R Y E L E C T R O N I C S I 1 3 . 2 - F O R T N E R Op-amps can be used as a current source all alone.

The current source only depends on Rf2 and Vout.

Iout+

Rf1

I1

I2

Vin

Ri2

Rf2Ri1

Rf2Rf1--------

Ri2Ri1--------=

IoutVinRf2--------=10 of 10

Differentiator and IntegratorDifferentiator An amplifier can utilize the relation between charge and current. The current is converted to a voltage For a sinusoidal input vin = V0sinwt, The amplitude increases with increasing frequency

Integrator The input current can be converted into a charge Solving for vout: With a sine wave input The amplitude decreases with frequency.

Stabilized Integrator The constant term from the integral is undesirable on the output. A switch can be used to discharge the capacitor. The circuit can provide a path for the capacitor to discharge.The feedback resistor and capacitor have a parallel impedanceThe break frequency is f = 1/2pRC = 0.007 Hz .Only DC can feedback for amplification.

Limited Differentiator Differentiator has large amplification at high frequencies. A high frequency cutoff above the signal range is often needed. Combine an integrator and differentiator. The differentiator has a low cutoff The integrator begins working at The differentiator will perform well between those two frequencies.

DC AmplifiersLogarithmic Amplifier Use a transistor in the feedback network of an inverting amplifier. RB compensates for the bias current. The current Iin is given by This current must flow into the collector of the transistor The base-emitter voltage is equal to the negative of Vout The output depends on the logarithm of the input voltage.

Analog Product Combine logarithm amplifiers and summation amplifiers The final stage is an exponentiator to return the product of the inputs.

Current-to-Voltage Converter Operational feedback relies on current.This is used for small input signals such as photonic devices.A parallel feedback capacitor can be used to reduce high frequencies.The output is usually measured in V/mA. The op-amp provides low output impedance, hence higher power.

Current-to-Current Converter An extra resistor to defines the current.From Kirchoffs Laws:Combine the equations:

Voltage-to-Current Converter An op-amp follower can be used to drive a conventional transistor current source. The current Iout splits through the FET and BJT. No current passes through the gate of the FET or the v- op-amp input. All current Iout is present through the resistor IR. VE = Vin from the op-amp voltage rule. Op-amps can be used as a current source all alone. The current source only depends on Rf2 and Vout.